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# At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second?

Solution

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First, we convert the linear speed units to meters per second. Then, we proceed to calculate the angular velocity with the given diameter $\left(d\right)$. Note that we have substituted the radius $\left(r\right)$ for $d/2$.

\begin{align*} v&=500 \text{ } \dfrac{\text{km}}{\text{h}} \cdot \dfrac{1000 \text{ m}}{1 \text{ km}} \cdot \dfrac{1 \text{ h}}{3600\text{ s}} \\ &=139 \text{ } \dfrac{\text{m}}{\text{s}} \\ \omega&=\dfrac{v}{r} \\ &=\dfrac{2v}{d} \\ &=\dfrac{2\left(139\right)}{60.0} \\ &=4.63 \text{ } \dfrac{\text{rad}}{\text{s}} \\ &=4.63 \text{ } \dfrac{\text{rad}}{\text{s}} \cdot \dfrac{1 \text{ rev}}{2\pi \text{ rad}} \\ &=0.737 \text{ } \dfrac{\text{rev}}{\text{s}} \end{align*}

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